Algebraic Fractions: Eduqas GCSE Higher June 2022 Paper 2 Q23

Higher June 2022 Paper 2 Q23

23. Debra buys a total of 4 kg of flour and butter at a market.
She pays £7 for her flour and £9 for her butter.

She buys \(x\) kg of flour.

(a) Complete this table. [1]

	Total cost (£)	Number of kg bought	Cost per kg (£)
Flour	7	\(x\)	\(\dfrac{7}{x}\)
Butter	9	\(4 - x\)

(b)

The cost per kilogram of her butter is £5 more than the cost per kilogram of her flour.

Use this information and the table in part (a) to show that

\[5x^2 - 4x - 28 = 0.\][5]

Scheme	Marks	Comments
\(\dfrac{9}{4 - x}\)	B1	Allow \(\dfrac{7}{x} + 5\)

Mark scheme (b)

Scheme	Marks	Comments
\(\dfrac{9}{4 - x} = \dfrac{7}{x} + 5\) oe	M1	FT ‘their \(\dfrac{9}{4 - x}\)’ provided of the form \(\dfrac{a}{bx + c}\) Equates correct or correct FT expressions
\(9x = 7(4 - x) + 5x(4 - x)\) oe	m1	Clears fractions e.g. \(9x = (7 + 5x)(4 - x)\); may be in stages
\(9x = 28 - 7x + 20x - 5x^2\) oe	m1	Multiplies out
\(9x = 28 + 13x - 5x^2\) oe	m1	Collects terms on RHS oe
Correct completion to given answer \(5x^2 - 4x - 28 = 0\)	A1	Not from wrong working

Mark scheme (c)

Scheme	Marks	Comments
\((5x - 14)(x + 2)\)	M2	M1 for \((5x \ldots 14)(x \ldots 2)\) Must be seen
\(x = 2.8\) (or \(x = -2\))	A1	CAO Allow if working solution seen in (b) and not contradicted in (c). Using trial and improvement Award B3 for a method leading to both solutions, namely x = –2 AND x = 2·8, otherwise B0.
\(\left(\dfrac{7}{2.8} =\right)\) (£)2.5(0)	B1	FT ‘their derived positive value of \(x\), providing two solutions with one positive and one negative si Allow this mark if the quadratic has been solved on the calculator (M0 A0 previously awarded).

Scheme	Marks	Comments
\(x = \dfrac{-(-4) \pm \sqrt{(-4)^2 - 4 \times 5 \times -28}}{2 \times 5}\) oe	M1	Must be seen; allow one slip in substitution for M1 m0 A0; formula must not be clearly incorrect. If the –(–4) not clearly included in the fraction then M0 unless corrected later. Award M1 m0 A0 for missing brackets on the -4² unless corrected but allow use of 4². Allow attempt to complete the square with at most one slip
\(x = \dfrac{4 + \sqrt{576}}{10}\) (or \(x = \dfrac{4 - \sqrt{576}}{10}\)) oe	m1
\(x = 2.8\) (or \(x = -2\))	A1	CAO Allow is working solution seen in (b) and not contradicted in (c)
\(\left(\dfrac{7}{2.8} =\right)\) (£)2.5(0)	B1	FT ‘their derived positive value of \(x\)’, providing two solutions with one positive and one negative si Allow this mark if the quadratic has been solved on the calculator (M0 A0 previously awarded)

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